January 02, 2012

TOPIC 2: Oxidative Addition & Reductive Elimination

Writing about web page http://www.warwick.ac.uk/go/ch3a2

Please click on “Comments” and type a question. I will respond as soon as I can

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  1. anonymous

    In the answer for topic 2, question 3 when you say “nucleophilic SNAr type mechanism” do you mean the Pt adding to the HBrC= and a negative charge going into the ring, before eliminating Br?

    14 Oct 2012, 13:08

  2. Peter Scott

    Yes, exactly. I have updated the answer to include the SNAr type mechanism and have also included this below.

    A nucleophilic SNAr type mechanism would give rise to a mixture of E and Z alkene products B and not the exclusive E observed. Also it will not give so much if any of the cisPt product as the combination step (the addition step) will give mostly transPt.

    14 Oct 2012, 14:31

  3. Richard Belmega

    For the 3-centre OA mechanism in question 3, why is 3-centre TS between the C-Br and the Metal? I thought that the C-Br bond is fairly polar and therefore this contradicts the need for ‘Low polarity X-Y bonds’ as stated in the notes. I mistakenly drew the 3-centre TS between the C=C and the metal due to it being an unsaturated organic electrophile. Could you please clarify this?



    17 Feb 2013, 19:01

  4. Peter Scott

    Yes, that is a very realistic idea, and I considered this. I thought that in the TS you suggest the alkene has to move away from the metal in order to move towards the product. In other words, theTS I drew is more product-like. That does not mean it is correct of course, but the other reason I decided against drawing an alkene coordination is that the TS/intermediate would be 18e, which is unusual for Pt(O), but not impossible. I agree with your polarity argument, but of course the presence of the double bond delocalises the system and makes it inherently less polar. The truth is likely to be somewhere in between my version and yours. What we do know from the data is that it must be a cis type addition, not classically two-step nucleophilic OA. The reason it goes this unusual way is because of the double bond.

    18 Feb 2013, 12:20

  5. Batman

    In a reductive elimination reaction is it promoted by bulky ligands to stabilise the lower coordination sphere and also electron withdrawing groups which will help to withdraw the electronic charge donated by the groups being eliminated?

    27 Feb 2013, 12:20

  6. Peter Scott

    That’s a simplistic treatment Batman but basically correct. RE has two interconnected requirements – reduction and elimination!

    Elimination: Bulky ligands promote dissociation to give lower CN species which (for reasons given in topic 2 slides 11-14 and also exemplified in topic 4 slides 7 and 12) are better able to undergo cis (reductive) elimination.

    Reduction: Metal centres are easier to reduce if they are in higher oxidation states or if they are cationic, or if they have electron withdrawing ligands. Hence, all else being equal, RE will be more favourable in these systems.

    27 Feb 2013, 12:24

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